Question

Use Maclaurin series to evaluate the limits. . \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}\)

Short answer

The limit is \( \frac{1}{2} \).

Step-by-step solution

Write the Maclaurin series of \( cos(x) \)

The Maclaurin series expansion for \( \cos(x) \) is given by: \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \] For small values of \( x \), higher-order terms become insignificant. Therefore, we can approximate \( \cos(x) \) as: \[ \cos(x) \approx 1 - \frac{x^2}{2} \]

Substitute the series into the limit expression

Replace \( \cos(x) \) in the given limit with its Maclaurin series approximation: \[ \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^2} \approx \lim_{x \rightarrow 0} \frac{1 - \(1 - \frac{x^2}{2} \)}{x^2} \]

Simplify the expression

Simplify the numerator: \[ 1 - \(1 - \frac{x^2}{2} \) = \frac{x^2}{2} \] The limit now becomes: \[ \lim_{x \rightarrow 0} \frac{\frac{x^2}{2}}{x^2} \]

Evaluate the limit

Simplify the fraction: \[ \lim_{x \rightarrow 0} \frac{\frac{x^2}{2}}{x^2} = \lim_{x \rightarrow 0} \frac{1}{2} = \frac{1}{2} \] Since \( x \) approaches 0, the fraction remains constant.

Key concepts

Maclaurin series

The Maclaurin series is a special case of the Taylor series expansion, where the expansion is made around zero. This series represents a function as an infinite sum of its derivatives at a single point. The general form of the Maclaurin series for a function \(f(x)\) is:
$$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f^{(4)}(0)x^4}{4!} + \thinspace \text{higher-order terms}$$

For example, the Maclaurin series of the cosine function is

$$\frac{\text{cos}(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \thinspace \text{higher-order-terms}}$$

This series converges very quickly for small values of \(x\), allowing us to use it for approximations in limit evaluations, as seen in this problem.

Limit evaluation

Limit evaluation is a key concept in calculus that describes the value that a function approaches as the input approaches a given point. The limit \( \frac{\text{lim}_{x \rightarrow 0}}\) means we need to find the behavior of the function as \(x\) gets closer and closer to 0.

For our exercise, \frac{1-\text{cos}(x)} {x^{2}}\(\thinspace, we used the Maclaurin series to approximate the cosine function, and then substituted it into the limit expression. By simplifying, we obtain \)\text{lim}_{x \rightarrow 0} \frac{\frac{x^2}{2}} {x^2} = \frac{1}{2}

Which shows how the quotient of the two functions behaves when \(x\) approaches 0.

Cosine function expansion

The cosine function has a well-known Maclaurin series expansion, which is very useful for approximating limits and solving various calculus problems. In general, for small values of x, the higher-order terms (third degree and higher) in the Maclaurin series become negligible. Therefore, we can approximate \( \text{cos}(x) \thinspace by \frac{\text{cos}(x) = 1 - \frac{x^2} {2}}

This approximation allows us to replace the cosine function in complex expressions, making them simpler to work with in calculus problems.

In our problem, this simplified form was substituted into \) \frac{1-\text{cos}(x)} {x^{2}}$, simplifying the limit evaluation.

Calculus

Calculus is a branch of mathematics that focuses on the study of change and motion. It deals heavily with limits, functions, derivatives, integrals, and infinite series.

The problem at hand demonstrates some of the fundamental tools of calculus: the limit and the series expansion. By understanding how to manipulate functions through their series representations and limits, complex problems can often be simplified and solved more easily.

In the specific exercise, we have combined knowledge of series (Maclaurin series) with the concept of limits to evaluate an otherwise challenging limit involving the cosine function. This showcases the power and utility of calculus in breaking down and solving intricate mathematical problems.